# Discuss how to evaluate $\oint_{C} \frac{e^{z^{2}}}{z^{2}}\,dz$ where C is a simple closed curve enclosing the origin

Discuss how to evaluate $\oint_{C} \frac{e^{z^{2}}}{z^{2}}\,dz$ where C is a simple closed curve enclosing the origin

Try using the fact that $e^{z}=\sum_{j=0}^\infty \frac{z_{j}}{j!}$ Could someone help me through this problem?

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### Method 1:

We obtain the Laurent series for $\frac{e^{z^2}}{z^2}$: $$\frac{1}{z^2}e^{z^2}=\frac{1}{z^2}(1+z^2+\frac{z^4}{2!}+...)=\frac{1}{z^2}+1+\frac{z^2}{2!}+...$$

This series shows that $\frac{e^{z^2}}{z^2}$ has a pole of second order at $z=0$ and the residue $Res_{z=0}=0$. Therefore: $$\oint_C \frac{e^{z^2}}{z^2}dz = 2\pi i\times 0=0$$

### Method 2:

The denominator has a zero of second order at $z=0$ (where the numerator does not have a zero at), therefore $\frac{e^{z^2}}{z^2}$ has a pole of second order at $z=0$. We know the residue of $f(z)$ at an $m$-th order pole equals to: $$Res_{z=z_0}=\frac{1}{(m-1)!}\lim_{z=z_0}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]$$ So: $$Res_{z=0}=\lim_{z=0}\frac{d}{dz}[z^2\frac{e^{z^2}}{z^2}]=\lim_{z=0}\frac{d}{dz}e^{z^2}=\lim_{z=0}2ze^{z^2}=0$$ From here we confirm the result we got in method 1.

### Method 3:

Cauchy's Integral Formula (The general form): $$\oint_C \frac{f(z)}{(z-z_0)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(z_0)$$

$e^{z^2}$ is analytic within $C$, Then: $$\oint_C \frac{e^{z^2}}{z^2}dz =2\pi i \frac{d}{dz}e^{z^2}|_{z=0}=2\pi i\times2ze^{z^2}|_{z=0}=0$$

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Since this has already been discussed in terms of residues, let's show another approach.

Since the only singularity is at the origin we can deform the contour to be centered at the origin with radius $1$. We assume the integral is to be taken in the counterclockwise direction. Let $z = e^{i\phi}$. Then $$\begin{eqnarray*} \oint_{C}dz\, \frac{e^{z^{2}}}{z^{2}} &=& i \int_0^{2\pi} d\phi \, e^{-i \phi} e^{e^{2i\phi}} \\ &=& i \int_0^{2\pi} d\phi \, e^{-i \phi} \sum_{k=0}^\infty \frac{1}{k!}\left(e^{2i\phi}\right)^k \\ &=& i \sum_{k=0}^\infty \frac{1}{k!} \int_0^{2\pi} d\phi\, e^{(2k-1)i\phi} \\ &=& 0 \end{eqnarray*}$$ since $\int_0^{2\pi} d\phi \, e^{(2 k-1) i \phi} = 0$ for $k\in \mathbb{Z}$. We can interchange the sum and integral since the series expansion for the exponential function on any bounded subset of $\mathbb{C}$ converges uniformly.

We can also show the integral over $\phi$ vanishes without using a series expansion, $$\begin{eqnarray*} I &=& i \int_0^{2\pi} d\phi \, e^{-i \phi} e^{e^{2i\phi}} \\ &=& i \int_{-\pi}^{\pi} d\phi \, e^{-i \phi} e^{e^{2i\phi}} \\ &=& -i \int_0^{2\pi} d\psi \, e^{-i \psi} e^{e^{2i\psi}} \\ &=& - I. \end{eqnarray*}$$ Therefore, the integral vanishes. Here we use the fact that the argument of the integral is periodic, with period $2\pi$, and then change variables, $\psi = \phi + \pi$.

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